Which statement would make this true |-27|_____|-45| A.greater B.equal C.less

Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n.

Vera_Pavlovna [14]

Split up the integration interval into 4 subintervals:

$\left[0,\dfrac\pi8\right],\left[\dfrac\pi8,\dfrac\pi4\right],\left[\dfrac\pi4,\dfrac{3\pi}8\right],\left[\dfrac{3\pi}8,\dfrac\pi2\right]$

The left and right endpoints of the $i$ -th subinterval, respectively, are

$\ell_i=\dfrac{i-1}4\left(\dfrac\pi2-0\right)=\dfrac{(i-1)\pi}8$

$r_i=\dfrac i4\left(\dfrac\pi2-0\right)=\dfrac{i\pi}8$

for $1\le i\le4$ , and the respective midpoints are

$m_i=\dfrac{\ell_i+r_i}2=\dfrac{(2i-1)\pi}8$

Trapezoidal rule

We approximate the (signed) area under the curve over each subinterval by

$T_i=\dfrac{f(\ell_i)+f(r_i)}2(\ell_i-r_i)$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4T_i\approx\boxed{3.038078}$

Midpoint rule

We approximate the area for each subinterval by

$M_i=f(m_i)(\ell_i-r_i)$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4M_i\approx\boxed{2.981137}$

Simpson's rule

We first interpolate the integrand over each subinterval by a quadratic polynomial $p_i(x)$ , where

$p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx$

It so happens that the integral of $p_i(x)$ reduces nicely to the form you're probably more familiar with,

$S_i=\displaystyle\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx=\frac{r_i-\ell_i}6(f(\ell_i)+4f(m_i)+f(r_i))$

Then the integral is approximately

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4S_i\approx\boxed{3.000117}$

Compare these to the actual value of the integral, 3. I've included plots of the approximations below.

3 0

3 years ago

Is the eye color of people on commercial aircraft flightseye color of people on commercial aircraft flights a discrete random v

const2013 [10]

Answer: The eye color of people on commercial aircraft flights is a discrete random variable.

Step-by-step explanation:

A discrete random variable is a variable with real values that are countable.

A discrete random variable has a probability that is between 0 and 1 for each possible values and the sum of all these probabilities equals 1.

7 0

3 years ago

Solve : x/3-2/5 = 2x/15-3/5

svetoff [14.1K]

Step-by-step explanation:

x/3-2/5=2x/15-3x

5x-6=2x-9

5x-2x=-9+6

3x=-3

3x/3=-3/3

x=-1

6 0

3 years ago

Consider the system of equations.

anzhelika [568]

Answer: You can multiply the top equation by -1 to eliminate the x variable.

And the solution is (2,4/3) in case you need it.

Step-by-step explanation:

2x + 3y = 8

2x + 6y = 12

If you multiply the upper equation or down equation by one, you will be able to eliminate the x variable.

-1( 2x + 3y) = -1(8) New equation: -2x -3y = -8.

Add the new equation you got by multiplying the top equation by -1 to the bottom equation.

Add them: -2x -3y = -8

2x + 6y = 12

3y = 4

y = 4/3

You can now input the value for y into the one of the equations and solve for x.

-2x - 3(4/3) = -8

-2x -4 = -8

+4 +4

-2x = -4

x = 2

8 0

3 years ago

31=x-14 gshshehheueuueheheueueu

fgiga [73]

Answer:

x=45

wasdrfgbhjnik

5 0

4 years ago

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